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2 years ago

Calculate the molarity of the two solutions.

Chemistry

1 answer:

zmey [24]2 years ago

4 0

Answer:

a) 0.100 M

b) 0.395 M

Explanation:

a) Calculate the molarity of a solution that contains 0.200 moles of NaOH (solute) in 2.00 L of solution

We will use the following expression for molarity.

[NaOH] = moles of solute / liters of solution

[NaOH] = 0.200 mol/2.00 L = 0.100 M

b) Calculate the molarity for a solution that contains 15.5 g of NaCl (solute, 58.44 g/mol) in 671 mL of solution

We will use the following expression for molarity.

[NaCl] = mass of solute / molar mass of solute × liters of solution

[NaCl] = 15.5 g / 58.44 g/mol × 0.671 L = 0.395 M

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The normal freezing point of a certain liquid

slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$

3 0

3 years ago

Molar mass c3h8

maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

3. 8.08 g

4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

2. 12.01×3= 36.03

3. 1.01×8= 8.08

4.(36.03/44.11)×100= 81.7%

5. (8.08/44.11)×100= 18.3%

8 0

2 years ago

How Do Glow Sticks Glow?

Gekata [30.6K]

Answer:

All liquid glow products depend on a chemical process known as CHEMILUMINESCENCE to produce their light. Chemiluminesence is a chemical reaction that causes a release of energy in the form of light. To produce this light the electrons in the chemicals become excited and rise to a higher energy level.

To utilise this process glowsticks contain two liquids; hydrogen peroxide and tert-butyl alcohol. When mixed together it is these liquids that create the glow. Fluorescent dyes are also used in the alcohol to alter the colour of the light emitted.

Explanation:

8 0

3 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

8 0

3 years ago

If 26.2 grams of a pure compound contain 8.77 × 1022 molecules, what is the molecular weight of this compound? Answer in units o

Nataly [62]

Answer:

Mw = 179.845 g/mol

Explanation:

Mw [=] g/mol

∴ w = 26.2 g

∴ 1 mol = 6.02 E23 molecules.......Avogadro's number

⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol

⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol

3 0

3 years ago